A250125
Coordination sequence of point of type 3.4.3.12 in 4-uniform tiling {3.3.4.3.4; 3.3.4.12; 3.3.12.4; 3.4.3.12}.
Original entry on oeis.org
1, 4, 6, 11, 13, 15, 23, 23, 33, 30, 33, 42, 41, 54, 46, 54, 58, 58, 73, 64, 75, 74, 79, 89, 81, 94, 92, 100, 105, 102, 110, 109, 119, 123, 123, 126, 130, 135, 140, 142, 144, 151, 151, 161, 158, 161, 170, 169, 182, 174, 182, 186, 186, 201, 192, 203, 202, 207, 217
Offset: 0
- Branko Grünbaum and G. C. Shephard, Tilings and Patterns. W. H. Freeman, New York, 1987.
- Joseph Myers, Table of n, a(n) for n = 0..1000
- Robert Connelly, Jeffrey D. Shen, Alexander D. Smith, Ball Packings with Periodic Constraints, arXiv:1301.0664 [math.MG], 2013.
- Robert Connelly, Jeffrey D. Shen, Alexander D. Smith, Ball Packings with Periodic Constraints, Discrete Comput. Geom. 52 (2014), no. 4, 754--779. MR3279548.
- Brian Galebach, Tiling 132 (in list of 4-uniform tilings).
- Brian Galebach, k-uniform tilings (k <= 6) and their A-numbers
- N. J. A. Sloane, A portion of the 3-uniform tiling {3.3.4.3.4; 3.3.4.12; 3.3.12.4; 3.4.3.12}. The four black dots labeled P,Q,R,S show the four types of point. The present sequence is for a point of type S.
- N. J. A. Sloane, Shows layers a(0)-a(6)
A250126
Coordination sequence of point of type 3.3.4.12 in 4-uniform tiling {3.3.4.3.4; 3.3.4.12; 3.3.12.4; 3.4.3.12}.
Original entry on oeis.org
1, 4, 9, 9, 12, 19, 21, 28, 27, 31, 38, 40, 48, 44, 49, 56, 57, 67, 63, 69, 73, 75, 85, 80, 88, 92, 95, 102, 98, 106, 109, 114, 121, 118, 123, 127, 132, 138, 137, 142, 147, 149, 156, 155, 159, 166, 168, 176, 172, 177, 184, 185, 195, 191, 197, 201, 203, 213, 208
Offset: 0
- Branko Grünbaum and G. C. Shephard, Tilings and Patterns. W. H. Freeman, New York, 1987.
- Joseph Myers, Table of n, a(n) for n = 0..1000
- Robert Connelly, Jeffrey D. Shen, Alexander D. Smith, Ball Packings with Periodic Constraints, arXiv:1301.0664 [math.MG], 2013.
- Robert Connelly, Jeffrey D. Shen, Alexander D. Smith, Ball Packings with Periodic Constraints, Discrete Comput. Geom. 52 (2014), no. 4, 754--779. MR3279548.
- Brian Galebach, Tiling 132 (in list of 4-uniform tilings).
- Brian Galebach, k-uniform tilings (k <= 6) and their A-numbers
- N. J. A. Sloane, A portion of the 3-uniform tiling {3.3.4.3.4; 3.3.4.12; 3.3.12.4; 3.4.3.12}. The four black dots labeled P,Q,R,S show the four types of point. The present sequence is for a point of type Q.
- N. J. A. Sloane, Shows layers a(0)-a(6)
A071910
a(n) = t(n)*t(n+1)*t(n+2), where t() are the triangular numbers.
Original entry on oeis.org
0, 18, 180, 900, 3150, 8820, 21168, 45360, 89100, 163350, 283140, 468468, 745290, 1146600, 1713600, 2496960, 3558168, 4970970, 6822900, 9216900, 12273030, 16130268, 20948400, 26910000, 34222500, 43120350, 53867268, 66758580, 82123650, 100328400, 121777920
Offset: 0
Cf.
A006542, (first differences of a(n) /18)
A006414, (second differences of a(n) /18)
A006322, (third differences of a(n) /18)
A004068, (fourth differences of a(n) /18)
A005891, (fifth differences of a(n) /18)
A008706.
-
Join[{0},Times@@@Partition[Accumulate[Range[40]],3,1]] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,18,180,900,3150,8820,21168},40] (* Harvey P. Dale, Aug 08 2025 *)
-
t(n) = n*(n+1)/2;
a(n) = t(n)*t(n+1)*t(n+2); \\ Michel Marcus, Oct 21 2015
A086460
Square array read by antidiagonals: T(n,k)=nk+0^n.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 4, 3, 0, 1, 4, 6, 6, 4, 0, 1, 5, 8, 9, 8, 5, 0, 1, 6, 10, 12, 12, 10, 6, 0, 1, 7, 12, 15, 16, 15, 12, 7, 0, 1, 8, 14, 18, 20, 20, 18, 14, 8, 0, 1, 9, 16, 21, 24, 25, 24, 21, 16, 9, 0, 1, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 0, 1, 11, 20, 27, 32, 35, 36
Offset: 0
Rows begin
1 0 0 0 0 ...
1 1 2 3 4 ...
1 2 4 6 8 ...
1 3 6 9 12 ...
1 4 8 12 16 ...
A086461
Symmetric version of square array A086460.
Original entry on oeis.org
1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 1, 1, 4, 6, 6, 4, 1, 1, 5, 8, 9, 8, 5, 1, 1, 6, 10, 12, 12, 10, 6, 1, 1, 7, 12, 15, 16, 15, 12, 7, 1, 1, 8, 14, 18, 20, 20, 18, 14, 8, 1, 1, 9, 16, 21, 24, 25, 24, 21, 16, 9, 1, 1, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 1, 1, 11, 20, 27, 32, 35, 36
Offset: 0
Rows begin
1 1 1 1 1 ...
1 1 2 3 4 ...
1 2 4 6 8 ...
1 3 6 9 12 ...
1 4 8 12 16 ...
As a triangle:
{1},
{1, 1},
{1, 1, 1},
{1, 2, 2, 1},
{1, 3, 4, 3, 1},
{1, 4, 6, 6, 4, 1},
{1, 5, 8, 9, 8, 5, 1},
{1, 6, 10, 12, 12, 10, 6, 1},
{1, 7, 12, 15, 16, 15, 12, 7, 1},
{1, 8, 14, 18, 20, 20, 18, 14, 8, 1},
{1, 9, 16, 21, 24, 25, 24, 21, 16, 9, 1}
-
t[n_, m_] = If[ n == 0 || n == m || m == 0, 1, n - m]*If[n == m || n == 0 || m == 0, 1, m]; Table[t[n, m], {n, 0, 11}, {m, 0, n}] // Flatten (* Roger L. Bagula, Sep 06 2008 *)
Original entry on oeis.org
6, 30, 60, 180, 210, 2310, 4620, 60060, 510510, 10810800, 116396280, 200560490130, 401120980260
Offset: 1
A024365 begins {6, 30, 60, 84, 180, 210, 210, 330, 504, 546, 630, 840, 924, 990, 1224, 1320, 1386, 1560, 1710, 1716, 2310, ...}.
A129912 begins {1, 2, 6, 12, 30, 60, 180, 210, 360, 420, 1260, 2310, 2520, ...}.
So, common entries encountered are {6, 30, 60, 180, 210, 2310, ...}.
Specifically, we see that A024365(1) = A129912(3), A024365(2) = A129912(5), A024365(3) = A129912(6), A024365(5) = A129912(7).
These are then the first four entries of the sequence (6, 30, 60, 180).
-
s = 6 Take[Sort[(Times @@ #)/12 & /@ ({Times @@ #, (Last[#]^2 - First[#]^2)/2} & /@ Select[Subsets[Range[1, 3600, 2], {2}], GCD @@ # == 1 &])], 1800]; f[m_] := f[m] = Union[Times @@@ Subsets[FoldList[Times, 1, Prime[Range[m]]]]][[1 ;; 100]]; f[10]; f[m = 11]; While[f[m] != f[m - 1], m++]; t = f[m]; Intersection[s, t] (* Michael De Vlieger, Oct 22 2015, after Harvey P. Dale at A020885 and Jean-François Alcover at A129912 *) (* or *)
ok[n_] := Block[{a, f = Power @@@ FactorInteger[2 n]}, SelectFirst[ Subsets[f, {1, Floor[ Length[f]/2]}], (a = Times @@ #; IntegerQ@ Sqrt[a^2 + (2 n/a)^2]) &, {}] != {}]; pr[n_] := Product[ Prime[n+1-i]^i, {i, n}]; upto[mx_] := Block[{ric, j=1}, ric[n_, ip_, ex_] := If[n < mx, Block[{p = Prime[ip + 1]}, If[ex == 1 && ok[n], Sow@ n]; ric[n p^ex, ip + 1, ex]; If[ex > 1, ric[n p^(ex - 1), ip+1, ex-1]]]]; Sort@ Reap[ While[pr[j] < mx, ric[2^j, 1, j]; j++]][[2, 1]]]; upto[10^12] (* much faster, Giovanni Resta, Mar 31 2017 *)
-
\\note: code does not generate the sequence, just checks for a matching PPT entry
genit(area)={myMax=floor(sqrt(2*area));i5=myMax;endless=0;soln=List();
while(i5>=2,dun=0;j=2.*myVal/i5; k=floor(j); if(j>k, dun=1 );if(dun<1,
c=sqrt(i5^2 + k^2);w=floor(c);if(c>w,dun=1); if(dun<1,if(gcd(k,i5)>1,dun=1 ));
if(dun<1,listput(soln,k); listput(soln,i5);listput(soln,w);listsort(soln);
print("soln a,b,c = ", soln[1]," ",soln[2]," ",soln[3] );dun=2;break ));
i5--;endless++);if(i5<=2&&dun<1,print("no solution ") );if(i5>2&&dun<2,
print("max iteration limit was hit ",endless) );print (endless);}
(C++)
#include
#include
using namespace std;
int main(){ifstream fin1,fin2;
int myValue,myValue2,ptr,fptr,i5,j5;
unsigned long list1[9999]={0};
unsigned long list2[999]={0};
unsigned long final[31]={0};
fin1.open("A024365.txt"); fin2.open("A129912.txt");
ptr=1;
while(ptr<9999)
{fin1>> myValue;fin1.get();list1[ptr]=myValue;
if(ptr<999)
{fin2>> myValue2;fin2.get();list2[ptr]=myValue2;}
ptr++;}
fin1.close();fin2.close();fptr=1;
for(i5=1;i5<9990;i5++)
{for(j5=1;j5<999;j5++){
if(list1[i5]==list2[j5] )
{
fptr++;
if(fptr>30){break;}
final[fptr]=list1[i5];
cout << final[fptr] << ",";
break;
}}if(fptr>30){break;}}}
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